If the mean free path of atoms is doubled then the pressure of the gas will become:
1. \(P/4\)           
2. \(P/2\)           
3. \(P/8\)           
4. \(P\)

The mean free path of molecules of a gas, (radius r) is inversely proportional to :
(1) r³

(2) r²

(3) r

(4) √r

The mean free path of gas molecules depends on (d = molecular diameter)

1. $\mathrm{d}$                                         

2. ${\mathrm{d}}^2$

3. ${\mathrm{d}}^{-2}$                                   

4. ${\mathrm{d}}^{-1}$

If the pressure in a closed vessel is reduced by drawing out some gas, the mean free path of the molecules:

Assertion: At a particular temperature, the value of the mean free path increases with a decrease in pressure.

Reason: All the gas molecules at a particular temperature possess the same speed.

When the gas in an open container is heated, the mean free path:

A closed container having an ideal gas is heated gradually to increase the temperature by 20%  The mean free path will become/remain:

1. 20% more

2. Same

3. 20% less

4. 33% less

Two ideal gases have the same number of molecules per unit volume and the radii of their molecules are \(r\) and \(3r\) respectively. The ratio of their mean free path in identical containers will be:
1. \(3:1\)
2. \(9:1\)
3. \(1:1\)
4. \(1:4\)

The mean free path for a gas, with molecular diameter \(d\) and number density \(n,\) can be expressed as:
1. \( \frac{1}{\sqrt{2} n \pi \mathrm{d}^2} \)
2. \( \frac{1}{\sqrt{2} n^2 \pi \mathrm{d}^2} \)
3. \(\frac{1}{\sqrt{2} n^2 \pi^2 d^2} \)
4. \( \frac{1}{\sqrt{2} n \pi \mathrm{d}}\)

On increasing number density for a gas in a vessel, the mean free path of a gas:

(1) Decreases

(2) Increases

(3) Remains same

(4) Becomes double